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Vector partial derivative

  1. Mar 8, 2009 #1
    1. The problem statement, all variables and given/known data
    A function f(x) : Rn ->R is said to be differentiable at point [tex]\vec{a}[/tex] provided that there exists a constant vector [tex]\vec{c} = (c_1, ... , c_n)[/tex] such that

    [tex]lim_(\vec{h} -> 0) \frac{f(\vec{a}+\vec{h}) - f(\vec{a}) - \vec{c}*\vec{h}}{||\vec{h}||}[/tex]

    Prove that if the multivariable function f(x) (here [tex]x = x_1, ..., x_n)[/tex] is differentiable at [tex]a = (a_1, ..., a_n)[/tex] then its first order partial derivatives at a exist.
    2. Relevant equations
    I know that the partial derivative definition is 342548949e92e400707a6864cb81bb00.png


    3. The attempt at a solution

    I've tried a few things but I've encountered a road block of sorts. I think what I have to do is provide the vector c such that the given equation somehow turns into the definition of the first order partial derivative. That means that instead of having f(a1+h1, a2+h2, ...) I need to make all the h's 0 except one, [tex]h_i[/tex] so the term would turn into [tex]f(a1, a2,..., a_i+h, a_{i+1}...)[/tex] But I'm not sure how to do that with the vector c... I may be way off base...

    Thank you for any hints/advice.
     
  2. jcsd
  3. Mar 8, 2009 #2

    Dick

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    You want to put [itex]\vec h[/itex] equal to [itex]h \vec {e_i}[/itex]. Where e_i is the ith basis vector and take the limit as the real number h approaches 0. Does that help? You can't pick c. That's a given. You can pick a particular form of h.
     
    Last edited: Mar 8, 2009
  4. Mar 8, 2009 #3
    Thank you for the response. What exactly is an "ith basis vector" though?
     
  5. Mar 8, 2009 #4

    Dick

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    e_i=(0,0,0,...,1,..0,0,0) with the 1 in position i. The same i as in your problem setup.
     
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